A) \[\sqrt{2}\]
B) \[\sqrt{2}\,(\sqrt{2}-1)\]
C) 2
D) none of these
Correct Answer: A
Solution :
\[l=\int\limits_{\pi /4}^{\pi /2}{\sqrt{2+\sqrt{2+2\cos 4x}}}\,\,dx=\int\limits_{\pi /4}^{\pi /2}{\sqrt{2-2\cos 2x}\,\,dx}\] |
\[=2\int\limits_{\pi /4}^{\pi /2}{\sin x\,\,dx}\] \[=2\,(-\cos x)_{\pi /4}^{\pi /2}=\sqrt{2}\] |
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